u∈Σn
| M
[1,2](u)|
Let us denote max
u∈Σn|M[1,k](u)|asA(Σ, n, k). We already know the values ofA(Σ, n, k) fork = 1 andk≥ bn2c+ 1. We investigate the value ofA(Σ, nk) for the intermediate values of k. In this section we consider the case k = 2. Using Theorem 3.2 we can calculate the |M[1,2]| but to find A(Σ, n,2) we need to consider every u in Σn. We shall use the constraints given by Rauzy multigraph to take into consideration every possible u in Σ = {0,1} to give a lower bound on A(Σ, n,2) for any Σ.But before discussing that we shall prove the following combinatorial result which uses the maximum value of binomial coefficient and its conditions.
Theorem 3.5. maxn(a+b)!
a!b!
(b+d)!
b!d! |a+ 2b+d= 4m and a, b, d, m∈N∪ {0}o is in Θ 2nn
and maximum value occurs whena =b =d=m.
Proof. Let the maximum value occurs for some n1, n2 ∈N∪ {0}such thata+b =n1 and b+d =n2. Now, h1 = max
(a+b)!
a!b!
a+b=n1
is attained when a=b if n1 is even or a, b∈ {bn21c,bn21c+ 1} with a6=b when n1 is odd.
Similarly, h2 = max
(b,d)!
b!d!
b+d=n2
is attained when b = d if n2 is even or b, d ∈ {bn22c,bn22c+ 1} with b 6=d when n2 is odd. It is to be noted that n1 and n2 has to be simultaneously even or odd as n1+n2 is even.
Let bothn1andn2both are even. In that case to attain the maximum the relation n1 =n2 has to satisfy. Otherwise (a+b)!a!b! (b+d)!b!d! will be ≤h1h2 which is attained when a = b = d = m. Hence, maximum value of (a+b)!a!b! (b+d)!b!d! is h1h2 = (2m)!(2m)!(m!)4 and a=b=d=m.
Let, both n1 and n2 be odd. Without loss of generality let n1 < n2. Then to attain h1h2 we will have a =bn21c, b=bn21c+ 1 =bn22c, d=bn22c+ 1. Which means a=m−1, b=m, d=m+ 1. Then
h1h2 (2m)!(2m)!
(m!)4
=
(2m−1)!(2m+1)!
(m−1)!m!m!(m+1)!
(2m)!(2m)!
(m!)4
= 2m2+m 2m2+ 2m <1
Hence the maximum is attained only when a = b = d = m. Now from the asymptotic value of central binomial coefficient we have (2m)!m!m! is Θ(√4mm) [13].
We conclude that asymptotic value of (2m)!(2m)!(m!)4 is Θ(42mm ) which is Θ(2nn).
Theorem 3.6. A(Σ, n,2) is Ω(2nn) for all finite alphabetΣ with |Σ| ≥2.
Proof. As u ∈ {0,1}n the Rauzy multigraph G2u will have at most two nodes with label 0 and 1. Having only one node will lead us to the trivial case of unique reconstruction. LetG2u is given by the Figure3.4 wherea, b, c, d are multiplicities of the edges (0,0),(0,1),(1,0),(1,1) respectively and hence a+b+c+d=n−1.
0 1
a
b c
d
Figure 3.4: Rauzy multigrpah G2u
From Theorem 3.2 if p1 corresponds to the symbol 0 and p2 corresponds to the
symbol 1 then the matrix
L(P, Q,2) =
p1−a −b
−c p2−d
NowG2uhas to satisfy certain conditions to have an Eulerian path or Eulerian circuit.
For that we must have b =c or|b−c|= 1. There is no other restriction on a and d other than them being non negative integers.
Case b=c:
Without loss of generality, let the Eulerian walk starts from the node labelled by 0.
Applying Theorem 3.2, we have|u|0 =p1 =a+b+ 1 (asb=c) and|u|1 =p2 =b+d.
From Theorem 3.2, the number of reconstructions will be (p1−1)!(p2−1)!
a!b!b!d! det L(P, Q,2)
= (a+b)!
a!b!
(b+d−1)!
b!d! det
b+ 1 −b
−b b
= (a+b)!
a!b!
(b+d−1)!
(b−1)!d!
We have from Theorem 3.5 max
(a+b)!
a!b!
(b+d−1)!
(b−1)!d!
a, b, d∈N∪ {0} and a+ 2b+d =n−1
is in Θ 2n
n
Using symmetry we shall have the same result if we start the Eulerian walk from the node labelled as 1.
Case |b−c|= 1 :
Let b=c+ 1. In this scenario the eulerian walk has to start from the node labelled as 0. Then p1 =a+c+ 1 and p2 =d+c+ 1. The matrix
L(P, Q,2) =
p1−a −b
−c p2−d
=
c+ 1 −c−1
−c c+ 1
From Theorem 3.2 the number of reconstructions will be:
(p1−1)!(p2−1)!
a!(c+ 1)!c!d! det L(P, Q,2)
= (a+c)!
a!c!(c+ 1)
(c+d)!
c!d! det
c+ 1 −c−1
−c c+ 1
= (a+c)!
a!c!
(c+d)!
c!d!
Now similarly from Theorem 3.5 we have the maximum number of reconstructions in Θ 2nn
.
Thus, we have for any binary string u, the highest number of reconstruction from m[1,2](u) will be in Θ 2nn
. Naturally if we extend this result it to any arbitrary alphabet Σ with |Σ| ≥2 we have A(Σ, n,2) is Ω(2nn).
3.5 Finding max
u∈Σn
| M
[1,k](u)| for k = b
|u|2c
For any string u, let us express the reverse of u by ˜u. Intuitively as k increases, M[1,k](u) will decrease. Pi˜na and Uzc´ategui proved in [40] that if k ≥ b|u|2 c+ 1 then M[1,k](u) = 1. They have also characterized the strings in M[1,k](u) for k =b|u|2 c by the two following theorems:
Theorem 3.7. [40] Ifu, v ∈Σ2p and v ∈M[1,p](u) with u6=v for some alphabetΣ then u will be of the form rab˜r for some a, b∈Σ, r∈ {a, b}p−1 and v = ˜u.
Theorem 3.8. [40] If u, v ∈Σ2p+1 and v ∈M[1,p](u) with u6=v for some alphabet Σ then u and v will be of the form:
(i) u=wcandv = ˜wcwherew∈ {a, b}p witha, b, c∈Σandm[1,p](w) = m[1,p]( ˜w).
(ii) u=cwandv =cw˜wherew∈ {a, b}p witha, b, c∈Σandm[1,p](w) = m[1,p]( ˜w).
(iii) u=rabc˜r and v = ˜u where r∈ {a, b, c}p with a, b, c∈Σ.
(iv) u=rabcs and v =rcabs wherer, s∈ {a, b, c}p with a, b, c∈Σ.
(v) u=rabcs and v =rbcas wherer, s∈ {a, b, c}p with a, b, c∈Σ.
Lemma 3.9. [40] If u, v ∈ Σ2p+1 for some alphabet Σ and v ∈ M[1,p](u) with u = rabcs and v = rbca where r, s ∈ Σp−1 and a, b, c ∈ Σ then the following conditions hold:
(a) r, s∈ {a, b, c}p−1 (b) s1 ∈ {a, b}
(c) rk−i =c⇔si =b ∀i∈ {1, . . . , p−1}
(d) rk−i =a⇔si =a ∀i∈ {1, . . . , p−1}
(e) rk−i =c⇔rk−i−1 =b ∀i∈ {1, . . . , p−2}
(f) rk−i =b⇔rk−i−1 ∈ {a, c} ∀i∈ {1, . . . , p−2}
(g) rk−i =a⇔rk−i−1 ∈ {a, c} ∀i∈ {1, . . . , p−2}
From Theorem3.7it is clear that max
u∈Σ2p|M[1,p](u)|= 2 because mirror image or re- verse of a string is unique. In Theorem3.8the existence of ambiguous reconstruction of u from M[1,p](u) proves that max
u∈Σ2p+1|M[1,p](u)| ≥2.
We investigate if there can be a situation where max
u∈Σ2p+1|M[1,p](u)|>2.
In Theorem3.8 all the five different forms of the pair uand v actually expresses five different permutations v of the symbols of the string u such that m[1,k](u) = m[1,k](v) for k =b|u|2 c. Ifu and v is in the form (i) then let us sayv is permutation π1(u), if uand v is in the form (ii) then let us say v is permutation π2(u) and so on.
If v, x∈ M[1,k](u) and v = πj(u) and x 6=u and x 6= v then x 6=πj(u). So only 5 possible permutation of symbols in the string u are probable candidates to be a member of M[1,p](u).
Theorem 3.10. For p ≥ 2, max
u∈Σ2p+1|M[1,p](u)| = 3 and if |M[1,p](u)| = 3 then u is a binary string.
We shall be needing the following few results to prove Theorem3.10.
Lemma 3.11. Let x 6= u, u 6= v, x 6= v with v, x ∈ M[1,p](u) and v = π1(u) then x=π2(u) with all three strings being binary.
Proof. From Theorem 3.7 and the form (i) in Theorem 3.8, let u=q1Rq2abq2Rq˜ 1c
v =q1Rq2baq2Rq˜ 1c where q1, q2 ∈ {a, b} and R∈ {a, b}p−3 and a, b, c∈Σ.
If p = 2 then u will be of the form qabqc and v is in the form qbaqc where q∈ {a, b} and a, b, c∈Σ.
Case 1: (x=π2(u))
Sub case 1.1: Asxis in the form (ii) of Theorem3.8,xmust beq1Rq2aq2bRq˜ 1c by using Theorem3.7. But from Theorem 3.7 this would not be possible if c6∈ {a, b}. Here is an example with u, v and xin this form:
x=a aaaa ab aaaa u=aaaa ab aaaa a v =aaaa ba aaaa a
Sub case 1.2: Ifp= 2 then xhas to be of the form qbqac. By Theorem3.7, calso needs to be in {a, b} to have x in this form.
Case 2: (x=π3(u))
Sub case 2.1: Asxis in the form (iii) of Theorem3.8,xmust becq1Rq2baq2Rq˜ 1. Soq1Rq2 =cq1Rwhich meansq1 =q2 =cand henceu=x=cpbcp, which is not possible.
Sub case 2.2: Forp= 2, x=qqbac
Letq=a then u=x, which is not possible.
Letq=b then x6∈Mp(u), which is also not possible.
Case 3: (x=π4(u))
Sub case 3.1: Asxis in the form (iv) of Theorem3.8,xmust beq1Rq2q2abRq˜ 1c.
Letq2 =a. Then we have
u=q1 Raa ba Rq˜ 1c u=q1 Raa ab Rq˜ 1c
Clearly x ∈ π2(u) and this situation has already been considered in the previous case.
Now, let q2 =b. Then x=v, which is not possible.
Sub case 3.2: For p= 2, x=qqabc.
Letq =a then x6∈Mp(u) which is also not possible.
Letq =b then v =x, which is not possible.
Case 4: (x=π5(u))
Sub case 4.1: Asxis in the form (iv) of Theorem3.8,xmust beq1Rq2bq2aRq˜ 1c.
Letq2 =a then x=v, which is not possible.
Letq2 =b then x=π3(u) and this situation has already been considered in previous case.
Sub case 4.2: For p= 2, x=qbqac.
Letq =a then v =x, which is not possible.
Letq =b then x6∈Mp(u), which is also not possible.
Hence the only possible option for existence of u, v and x under given condition is x=π2(u) with all the three stings being binary.
Lemma 3.12. Let u, v and x are three distinct strings with v, x ∈ M[1,p](u) and v =π2(u). Thenx=π1(u) with all three strings being binary.
Proof. Cases where x = π1(u) are already taken care of in previous Lemma and x6=π2(u) as x6=v.
Case 1: (x=π3(u))
We have [m[1,p](u) =m[1,p](v)]⇔[m[1,p](˜u) = m[1,p](˜v)]. Thus [v, x∈M[1,p](u) and v =π2(u) and x=π3(u)]
if and only if
[˜v,x˜∈M[1,p](˜u) and ˜v =π1(˜u) and ˜x=π3(˜u)]
The later part is already taken care of and we have existence of different ˜u,˜v and ˜xmeans ˜v ∈π1(˜u),x˜∈π2(˜u) which impliesv ∈π2(u), x∈π1(u). Alsou, v and x are binary strings if and only if ˜u,˜v and ˜x are binary strings.
Case 2: (x=π4(u))
We have [m[1,p](u) =m[1,p](v)]⇔[m[1,p](˜u) = m[1,p](˜v)]. Thus [v, x∈M[1,p](u) and v =π2(u) and x=π4(u)]
if and only if
[˜v,x˜∈M[1,p](˜u) and ˜v =π1(˜u) and ˜x=π5(˜u)]
The later part has already been taken care of and we have existence of different
˜
u,v˜and ˜x means ˜v ∈π1(˜u),x˜∈π2(˜u) which impliesv ∈π2(u), x∈π1(u). Also u, v and x are binary strings if and only if ˜u,˜v and ˜x are binary strings.
Case 3: (x=π5(u))
We have [m[1,p](u) =m[1,p](v)]⇔[m[1,p](˜u) = m[1,p](˜v)]. Thus [v, x∈M[1,p](u) and v =π2(u) and x=π5(u)]
if and only if
[˜v,x˜∈M[1,p](˜u) and ˜v =π1(˜u) and ˜x=π4(˜u)]
The later part is already taken care of and we have existence of different ˜u,˜v and ˜xmeans ˜v ∈π1(˜u),x˜∈π2(˜u) which impliesv ∈π2(u), x∈π1(u). Alsou, v
and x are binary strings if and only if ˜u,v˜and ˜x are binary strings.
Lemma 3.13. Let u, v and x are three different strings with v, x ∈ M[1,p](u) and v = π3(u). Then v must be in the form of either π1(u) or π2(u) and x = π2(u) or x=π1(u) with all three strings being binary.
Proof. Cases wherex=π1(u) orx=π2(u) were already taken care of andx6=π3(u) as x6=v.
Case 1: (x=π4(u))
[v, x∈M[1,p](u) and v =π3(u) and x=π4(u)]
⇔[u, x∈M[1,p](v) andu=π3(v) and x=π2(v)]
⇔[˜u,x˜∈M[1,p](˜v) and ˜x=π1(˜v) and ˜u=π3(˜v)]
⇒[˜u,x˜∈M[1,p](˜v) and ˜x=π1(˜v) and ˜u=π2(˜v)], as u6=v, v 6=x, x6=u
⇔[u, x∈M[1,p](v) andu=π1(v) and x=π2(v)]
⇔[v, x∈M[1,p](u) and v =π1(u) and x=π3(u)]
⇒[v, x∈M[1,p](u) and v =π1(u) and x=π2(u)], as u6=v, v 6=x, x6=u
Case 2: (x=π5(u))
[v, x∈M[1,p](u) and v =π3(u) and x=π5(u)]
⇔[u, x∈M[1,p](v) and u=π3(v) and x=π1(v)]
⇒[u, x∈M[1,p](v) and u=π2(v) and x=π1(v)], as u6=v, v 6=x, x6=u
⇔[v, x∈M[1,p](u) and v =π2(u) and x=π4(u)]
⇒[v, x∈M[1,p](u) and v =π2(u) and x=π1(u)]
Lemma 3.14. Let u, v and x are three different strings with v, x ∈ M[1,p](u) and
u∈Σn M
v = π4(u). Then v must be in the form of either π1(u) or π2(u) and x = π2(u) or x=π1(u) with all three strings being binary.
Proof. Cases wherex=π1(u) orx=π2(u) orx=π3(u) are already taken care of and x6=π4(u) asx6=v. So x=π5(u). From Theorem 3.8, we have u=rabcs, v =rcabs and x = rbcas. Also we have x ∈ M[1,p](u) and x = π5(u). From Lemma 3.9, we get rp−i = c ⇔ si = b and rp−i = a ⇔ si = a ∀i = {1,2, . . . , p−1}. Here v ∈ M[1,p](x) and v = π5(x). So using Lemma 3.9, we have rp−i = a ⇔ si =c and rp−i =b⇔si =b ∀i={1, . . . , p−1}.
Let for some i, si = a. So, rp−i = a but that means si = a. Hence c = a and u, v, x are binary strings. Also ifc=a then x=π1(u) and v =π2(v).
Let for somei,si =b. So,rp−i =c. Also So,rp−i =b. Hence,b=candu, v, xare binary strings. Also if b =c then x= π3(u) and v =π1(u) which means x =π2(u) and v =π1(u) as u6=v, v 6=x and x6=u.
Let for some i, si = c. So, rp−i = a. Which means si = a. Hence c = a and u, v, x are binary strings. Also ifc=a then x=π1(u) and v =π2(v).
(Proof of Theorem 3.10) So, we have found that whenever we are choosing three different stringsu, v and xof length 2p+ 1 such thatm[1,p](u) = m[1,p](v) = m[1,p](x) then there is only two types of permutations of symbols ofuare allowed to bev and x. Also all the three strings needed to be from an alphabet of size 2 to be in those forms. Hence we have the proof of Theorem 3.10.